Friction problem with loop (work and energy)

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The discussion focuses on determining the minimum height required for an amusement park car to complete a loop with friction considered. The user successfully solved the problem without friction but struggles when incorporating it, particularly in calculating the work done by friction during the loop. They have derived equations for the car's velocity at various points but are confused about estimating the normal force and the work done by friction in a circular motion. The user seeks clarification on how to accurately estimate the work done by friction when the car has an initial velocity and is navigating the loop. The thread highlights the complexities of integrating forces and motion in a frictional context.
pulcroman
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Hello, i appreciate your help!

1. Homework Statement
In this website the drawing of the situation in the first exercise can be seen, for better understanding.
http://wwwprof.uniandes.edu.co/~gtel...s-semana10.pdf

The problem asks me to determine the minimum height h to allow an amousement park car to complete a loop with radius r without falling. The inicial velocity is 0. In the beginning, i had to do it without friction, with no problem. The horribleness started when i tried to do it with friction, more specifically in the loop part.
I use v1 for the speed after it falls the height h, v2 for the time it continues by the flat, and v3 for the speed in the tallest part of the loop. (Question in the last sentencie, for the impatient readers :) )

2. Homework Equations

E2-E1=-W; W=Fk dx; Fk=µk N. E(1,2)= Ek+Ep

3. The Attempt at a Solution

For the beginning of the movement, the fall from the height h to the floor, assuming is a straight plane is given by:
m g h - 1/2 m v1^2 = -µk m g cosø (h/senø)<---that last is the distance from h.
v1=√(2 g h-g h µk cotø) <---- i'll go fast, in he first steps. I just simplified v1.

for the second part, when the car moved a small part on a flat, i had
1/2 m v2^2 - 1/2 mv1^2=-µk m g cos ø x <--- x=dx, distance.
v2=√(2gh - g h µk cotø -2 g x µk)

Now, in the loop.
I've tried everything, and can't come up with a reliable solution. I know i can use
1/2 m v3^2+ mg(2r) - 1/2 m v2^2=-Fk x2

well, x2 is πr because it'll just half the circle, but I am having a real bad time finding the normal force for the Fk=µk N. Because, for a circular movement i got:
N= (m v^2)/r + m g cos ø
is v changing? if i integrate for ø from 0 to π it'll give 0! if i want to find the tangential acceleration i'll complicate myself even more. So, in a summary, the big question i want to ask is

How do you estimate the work made by friction on a loop with given inicial velocity, so the car won't fall? Thank you very much!
 
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The link does not work. Show a picture in an other way. ehild
 
ok, there it goes http://wwwprof.uniandes.edu.co/~gtellez/fisica1/ejercicios-semana10.pdf

i wanted to add: what will be the TOTAL work made by friction passing on half of a loop (this is for purposes of integration, if necessary !)
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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